# Time Value of Money — Interest Paid on Mortgage

You you bought a house worth \$328,000. You paid 25% of the purchase price in cash and arranged a 25 year mortgage with a rate of 4.0% compounded semi-annually for the remaining balance. The mortgage has an amortization period of 25 years. How much interest will you pay in the first 7 years (assuming that the first payment is made at the end of the first month)?

So far, I have that PV=\$328,000 * 0.75=\$246,000, r=0.00330589 (using effective rate formula: (1+r)^6=(1+0.04/2) ) and n=25 * 12=300. Using the present value of an ordinary annuity:

PV=PMT[(1-(1+r)^-n)/r]

I solved for PMT and got PMT=\$1294.009652 for the monthly payments. The number of payment periods still remaining after 7 years is 18*12=216. The PV of the outstanding balance (FV of 246,000 - FV of 84 PMTs) is \$199,539.6457. However, I don't really know what to do after that. The correct answer is \$62,236.46 but I don't know how they got that. How do I calculate the interest paid in the first 7 years?

• What is the background of this problem? I ask because I am unaware of mortgages that compound semi-annually. Typical is a calculation of interest and payment each month. Is this a homework problem? Dec 7, 2021 at 22:40
• Yes it is a homework problem for a class. I do apologize if this is the wrong place to ask this type of question. Dec 7, 2021 at 22:52
• @JTP-ApologisetoMonica Many fixed-rate mortgages in Canada have semi-annual compounding for the rate, even though payments are typically monthly. Dec 8, 2021 at 12:09

Edit:

Actually, since you already got the Outstandind Principal right after 84 months = \$199,539.6457, you knew that:

Total Principal Paid = 246,000 - 199,539.6457 = \$46,460.3543

Total Interest Paid = 84 x 1,294.009652 - 46,460.3543 = \$62,236.456468

Start with PMT and r. Your PMT = \$1294.009652 and r = 0.00330589 are correct.

Then use second formula at: https://en.wikipedia.org/wiki/Mortgage_calculator#Total_interest_paid_formula • P = 246,000
• r = 0.00330589
• c = PMT = 1294.009652
• N = 84

You will get \$62,236.46

Alternatively just use the Interest paid function of BA II PLUS: https://education.ti.com/download/en/ed-tech/ADF11FB65B284B6195B0A7E9502784BA/5DC3E70F3C8040E499D704B583646E1D/BA_II_PLUS_EN.pdf

You may also play with the relationship:

Total Interest Paid + Total Principal Paid = 84 x 1294.009652

• Very helpful answer (the edit especially)! Thank you very much for your help! Dec 8, 2021 at 21:07

With `s` as the loan amount

``````hse = 328000
s = hse (1 - 0.25) = 246000
``````

and `r` the monthly rate

``````i = 0.04
r = (1 + i/2)^(2/12) - 1 = 0.00330589

n = 25*12 = 300
``````

the payment amount `d` is

``````d = r (1 + 1/((1 + r)^n - 1)) s = 1294.01
``````

The principal balance in month `x` is given by `p(x)` (see link)

``````p(x) = (d + (1 + r)^x (r s - d))/r
``````

and the interest paid in month `x` is given by `int(x)`

``````int(x) = p(x - 1) r
= d + (1 + r)^(x - 1) (r s - d)
``````

The accumulated interest to month `x` is given by `interestsofar(x)`

(formula obtained by induction, summing `int(k)` from `k = 1` to `x`) ``````interestsofar(x) = (d - d (1 + r)^x - r s + r (1 + r)^x s + d r x)/r

interestsofar(7*12) = 62236.46
``````